Question: $g(n)=-72\cdot\left(\dfrac{1}{6}\right)^{{\,n-1}}$ Complete the recursive formula of $g(n)$. $g(1)=$
Explanation: From the explicit formula, ${-72}\cdot\left({\dfrac16}\right)^{n-1}$, we can tell that the first term of the sequence is ${-72}$ and the common ratio is ${\dfrac16}$. This is the recursive formula of the sequence: $\begin{cases} g(1)={-72} \\\\ g(n)=g(n-1)\cdot{\dfrac16} \end{cases}$